Question about $4$-manifolds and intersection forms

This is a question related to an earlier question of mine:

My current understanding is: we can turn a topological space $X$ into a module by taking any of its homologies and once we have a module we define a bilinear form on $H_i(X)$. For example, if we are taking the first homology we can define an bilinear form by defining $x \cdot y$ to be the intersection number of two representatives $x,y$ where $[x], [y] \in H_1 (X)$. Then this gives us a quadratic form $Q(x) = \frac{1}{2} x \cdot x \mod 2$. We can use this if we want to show that two topological spaces are not homeomorphic as follows: pick a symplectic basis $a_i, b_i$ for the space (we know that such a basis always exists). Then we define $A(Q) = \sum_i Q(a_i) Q(b_i)$. If this evaluates to $0$ in one space and to $1$ in the other we know that the two spaces are not homeomorphic.

Question 1: Is my current understanding correct? Or am I missing anything?

Question 2: I understand how to define a bilinear form if I take the first homology. But this might not lead anywhere because even though the two spaces might be non-homeomorphic, their first homology groups might still coincide. So I might want to take a higher homology. How do I define a bilinear form for the $k$-th homology and what is the geometric meaning?

Question 3: I found this Wikipedia article about intersection forms. Although I am quite sure that the "intersection form" is another bilinear form just like the intersection number for the first homology, I'm confused about why I could only find additional information about $4$-manifolds. What is special about $4$-manifolds? Am I wrong in assuming that I can endow any topological space with a bilinear form?

Question 4: We do this over $\mathbb Z / 2$ for convenience, right? Because it lets us ignore orientation. But we could also consider bilinear forms over any other field, is this correct?

Edit

Question 5: Poincaré duality gives us a way to define an intersection pairing without much nasty fiddling if the topological space $X$ is a closed oriented manifold. But what are the absolute minimal requirements on $X$ in order to be able to define an intersection form on it?

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There is at least a way to generalise the 4-dimensional intersection form to a general billinear form on $2n$-dimensional connected oriented manifolds – Juan S Aug 24 '12 at 2:24
Dear @JuanS, thank you for your comment! To $n$-manifold won't work? Why does it have to be $2n$? – Rudy the Reindeer Aug 24 '12 at 5:50
@Matt: the cup product gives a map $H^i \times H^j \to H^{i+j}$. If you want $i = j$ then you get a map $H^i \times H^i \to H^{2i}$. You want $H^{2i}$ to be the top cohomology so you can use Poincaré duality to identify it with $H_0$ and this requires that the manifold be $2i$-dimensional. – Qiaochu Yuan Aug 24 '12 at 6:07
@Matt: also, it's inaccurate to call these things inner products. An inner product is always positive-definite. – Qiaochu Yuan Aug 24 '12 at 6:07
Dear @QiaochuYuan, thank you for your helpful comments! I corrected the question. – Rudy the Reindeer Aug 24 '12 at 6:10

Q1: no. The intersection pairing is not defined on general topological spaces.